{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

04win-m1sols

# 04win-m1sols - SOLUTIONS TO MATH 51 MIDTERM 1 1 Find all...

This preview shows pages 1–3. Sign up to view the full content.

SOLUTIONS TO MATH 51 MIDTERM 1 January 29, 2004 1. Find all solutions of the following system: x 1 - x 2 + x 3 + 2 x 4 = 3 x 2 + x 3 + x 4 = 3 x 1 + x 2 + 3 x 3 + 4 x 4 = 9 Solution. Write the augmented matrix and then use Gaussian elimination: 1 - 1 1 2 3 0 1 1 1 3 1 1 3 4 9 1 - 1 1 2 3 0 1 1 1 3 0 2 2 2 6 1 0 2 3 6 0 1 1 1 3 0 0 0 0 0 So x 1 + 2 x 3 + 3 x 4 = 6 x 2 + x 3 + x 4 = 3 The free variables are x 3 and x 4 , so the solutions are: x 1 = 6 - 2 x 3 - 3 x 4 , x 2 = 3 - x 3 - x 4 , ( x 3 R , x 4 R ) 2. Let L be the intersection of the two planes x + y + z = 4 and 2 x + 3 y + z = 9 . Find a parametric equation for L . Solution. Write the augmented matrix and use Gaussian elimination: fl fl fl fl 1 1 1 4 2 3 1 9 fl fl fl fl fl fl fl fl 1 1 1 4 0 1 - 1 1 fl fl fl fl fl fl fl fl 1 0 2 3 0 1 - 1 1 fl fl fl fl . so x + 2 z = 3 and y - z = 1, or (moving the free variable z to the right hand side) x = 3 - 2 z and y = 1 + z . Thus the intersection is given by x y z = 3 - 2 z 1 + z z = 3 1 0 + z - 2 1 1 . 3(a) . Suppose u , v , and w are points in R n such that k u k = k v k = k w k = 1 and such that w = - u . Suppose also that v is not equal to u or to w . Prove that the triangle Δ uvw has a right angle at v . 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Solution. The vector from u to v is v - u . The vector from w to v is v - w . We want to show that these two vectors are orthogonal, so we calculate the dot product: ( v - u ) · ( v - w ) = ( v - u ) · ( v + u ) = v · v + v · u - u · v - u · u = k v k 2 -k u k 2 = 1 2 - 1 2 = 0 3(b) . Suppose x , y , and z are vectors in R n whose norms are 1, 2, and 3, respec- tively. Suppose each vector is orthogonal (i.e., perpendicular) to each of the other two. Find a scalar c such that the vector x + c y - z is orthogonal to the vector x + y + z . Solution The vectors x + y + z and x + c y - z vectors will be orthogonal provided their dot product is 0. When we multiply it out (i.e., use the distributive property), all the “mixed” terms ( x · y , x · z , etc.) are 0 by orthogonality. So 0 = ( x + y + z ) · ( x + c y - z ) = x · x + c y · y - z · z = k x k 2 + c k y k 2 -k z k 2 = 1+4 c - 9 = 4 c - 8 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 7

04win-m1sols - SOLUTIONS TO MATH 51 MIDTERM 1 1 Find all...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online